Why is the topology on $\operatorname{Proj} B$ induced from that on $\operatorname{Spec}(B)?$

Edit: As pointed out in the comments below, I was not using the correct definition of $I^h$. I do recall working with this construction some time ago: it is actually the largest homogeneous ideal contained in $I$. An interesting fact that may be useful is that, if $P$ is prime in $B$, then $P^h$ is a homogeneous prime ideal. If a solution to the actual question comes to me, I will of course let you know.

Old, incorrect answer:

This is clear once you realize that $I^h$ is the smallest homogeneous ideal containing $I$. Thus if $P$ is a homogeneous ideal, then $I \subseteq P \iff I^h \subseteq P$ (the reverse implication following from $I \subseteq I^h$). In my opinion, the equality follows easily from this.

To explicitly show that $V_+(I^h) \subseteq V(I)\cap\operatorname{Proj}(B)$ as you want, suppose that $P \in \operatorname{Proj}(B)$ with $I^h \subseteq P$. Then $I \subseteq I^h \subseteq P$, so $P \in V(I) \cap \operatorname{Proj}(B)$.

$$V(I)\cap\operatorname{Proj}(B)=V_+(I^h)$$

This is false. Consider $I=(X+1)$ in $B=K[X,Y]$. Then $I^h=(0)$, so $V_+(I^h)=\operatorname{Proj}B$. On the other side, we have $V(I)\cap\operatorname{Proj}B=\emptyset$.

So more importantly, if $f \in S$, write $f = f_{0} + f_{1} + \cdots + f_{n}$. Given $\mathfrak{p} \in \text{Proj}S$, we have $\mathfrak{p} \ni f$ if and only if $\mathfrak{p} \ni f_{0}, \cdots, f_{n}$. Thus $V(f) \cap \text{Proj}S = V_{+}(f_{0}) \cap \cdots \cap V_{+}(f_{n})$, and the topology is inherited!